Decomposition of ethane, \(\frac{d\left[\mathrm{C}_2 \mathrm{H}_6\right]}{d t}=k\left[\mathrm{C}_2 \mathrm{H}_6\right]\) proceeds through a complex mechanism, which includes 5 steps. The overall rate constant \((k)\) can be expressed as \(k=\frac{k_1 k_2 k_3}{k_2 k_5}\). where \(k_1, k_2, k_3, k_4, k_5\) are the rate constant of the 5 steps.
If the activation energies of each of the steps respectively are \(E_1=1 E, E_2=2 E\),

\(E_3=3 E, E_4=4 E, E_5=5 E\), where \(E=20 \mathrm{~kJ} / \mathrm{mol}\). Then find the overall activation energy of the decomposition.

1. A \(6.67 \mathrm{~kJ} / \mathrm{mol}\)
2. B \(3.33 \mathrm{~kJ} / \mathrm{mol}\)
3. C \(20 \mathrm{~kJ} / \mathrm{mol}\)
4. D \(10 \mathrm{~kJ} / \mathrm{mol}\)